Mersenne Primes
The Lucas-Lehmer Test. Let p be an odd prime, and define n Definition.Definition. A AMersenneMersenne prime primeisis a a prime prime of of the the form form 2 2nn 1.1. Definition. A Mersennerecursively prime is a prime of the form 2n 1. p 2 p Equivalently,Equivalently,...... ofof the the form form 2 2pp 11 where whereppisis a a prime. prime. •Equivalently, ... Lof0 the=4 formand 2p 1L wheren+1 =p Lisn a prime.2 (mod (2 1)) for n 0. •• MersenneMersenne primes primes are are related relatedp to toperfectperfect numbers numbers.. Euler Euler Mersenne primesThen are related2 to1 isperfect a prime numbers if and. Euler only if Lp 2 =0. ••• • showedshowedshowed that that that ((mm(m)=2)=2)=2mmm,,, where where wheremmmisisis even even even if if if and and and only only only if if if showedp that1 p (m)=2m,p where m is even if and only if mm=2=2pp 11(2(2pp 1)1) and and 2 2p 11 is is prime.p prime. m =2p 1(2p 1) where p and 2p 1 are primes. 5788516157885161 TheThe largest largest known known prime prime is is 2 25788516177232917 1.1. •The largest known prime is 257885161 1.1 •• O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas-LehmerThe Primality Lucas Primality Test Test
Fix integers P andTheQ Lucas-Lehmer.LetD = P 2 Test.4Q. DefineLet p recursivelybe an odd prime, and define un and vn by recursively u0 =0,u1L=1=4,un+1and= PuLn Qu=n L12for n2 (mod1, (2p 1)) for n 0. 0 n+1 n v0 =2,v1 = P, andp vn+1 = Pvn Qvn 1 for n 1. Then 2 1 is a prime if and only if Lp 2 =0. (p 1)/2 Maple’sIf p is an Version odd prime and p - PQ and D 1(modp), ⌘ then p u . Maple’s Version| p+1 Maple’sTake VersionQ =1. • Maple’s Version Idea: Given a large positive integer n,ifn isP prime,2 4 there is TakeFindQ first=1.P where the Jacobi symbol(p 1)/2 = 1. a• 50-50 chance that a D will satisfy D 1(modn). Take• Q =1. P 2 ⌘ n4 • Take Q =1. ✓2 ◆ PlayFindShould with firstP checkPandwhere ifQnuntil theis a Jacobi square you find symbol (recall such 1093 aP 2D andwith4 3511).=n -1.PQ. • Find• first P where the• Jacobi symbol n = 1. 2 ComputeFind firstunP+1 wherequickly the and Jacobi check symbol if n un✓+1. If not,◆ = then1.n is P 4 • Make use of the identities, where|n 1.n composite.Should• check If so, if thenn is a it square isFind likely (recall firstn isP prime. 1093✓where and the 3511).◆ Jacobi symbol = 1. • Should check if n is a• square (recall 1093 and 3511). n • Make use of the identities, where n 1. ✓ ◆ • Make use of the identities,Should where checkn 1. if n is a square (recall 1093 and 3511). HowMake do use we compute of the identities,u quickly? where n 1. •v = v2 2,v n=+1•v v P, Du =2v Pv 2n n 2n+1 Maken+1 n use of then identities,n+1 wheren n 1. Why does2 p up+1 if p is• a prime? v2n = v 2,v| 2n+1 = vn+1vn P, Dun =2vn+1 Pvn n2 v2n = vn 2,v2n+1 = vn+1vn P, Dun =2vn+1 Pvn Why2n shouldn we2 thinkn+1 n nis+1 likelyn a primen if n unn+1+1? n | v = v2 2,v = v v P, Du =2v Pv 2n n 2n+1 n+1 n n n+1 n O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas-LehmerThe Primality Lucas Primality Test Test
Fix integers P andTheQ Lucas-Lehmer.LetD = P 2 Test.4Q. DefineLet p recursivelybe an odd prime, and define un and vn by recursively u0 =0,u1L=1=4,un+1and= PuLn Qu=n L12for n2 (mod1, (2p 1)) for n 0. 0 n+1 n v0 =2,v1 = P, andp vn+1 = Pvn Qvn 1 for n 1. Then 2 1 is a prime if and only if Lp 2 =0. (p 1)/2 Maple’sIf p is an Version odd prime and p - PQ and D 1(modp), ⌘ then p u . Maple’s Version| p+1 Maple’sTake VersionQ =1. Suppose n 7(mod12)is • ⌘ Idea: Given a large positiveprime. integer We cann,if taken isPP prime,2 =4.4 there is TakeFindQ first=1.P where the Jacobi symbol(p 1)/2 = 1. a• 50-50Take• Q chance=1. that a D will satisfy D 2 n 1(mod n). • Take Q =1. P✓ ⌘ 4 ◆ •PlayFind with firstPPandwhereQ until the Jacobi you find symbol such a 2D with=n -1.PQ. 2 Should check if n is a squarev =4 (recall,v =14 1093P,... and4 3511). vn+1 = v n 2(n 0) • Find• first P where the Jacobi1 symbol2 n = 2 1. 2 •Compute un+1 quickly and check if n un✓+1. Ifn not,◆ then n is • ShouldMake check use of if then is identities, a square (recall where|n 1093✓ 1. andn 3511).◆ composite.• • If so, then it is likely n is prime.✓ ◆ Should check if n is a square (recall 1093 andL 3511).= v n • Make use of the identities, where n 1. n 2 • Make use of the identities, where n 1. •How do we2 compute un+1 quickly? v2n = vn 2,v2n+1 = vn+1p vn P, Dun =2vn+1 Pvn N =2 1 is a prime v(N +1)/4 is divisible by N Why does2 p up+1 if p is a prime? () v2n = v 2,v| 2n+1 = vn+1vn P, Dun =2vn+1 Pvn n2 v2n = vn 2,v2n+1 = vn+1vn P, Dun =2vn+1 Pvn Why2n shouldn we2 thinkn+1 n nis+1 likelyn a primen if n unn+1+1? n | O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas-LehmerThe Primality Lucas Primality Test Test
Fix integers P andTheQ Lucas-Lehmer.LetD = P 2 Test.4Q. DefineLet p recursivelybe an odd prime, and define un and vn by recursively
u0 =0,u1 =1,un+1 = Pun Qun 12for n 1, p L =4 and Ln = L 2 (mod (2 1)) for n 0. Maple’s Version 0 +1 n v0 =2,v1 = P, andp vn+1 = Pvn Qvn 1 for n 1. Then 2 1 is a prime if and only if Lp 2 =0. (p 1)/2 If p isTake an oddQ =1. prime and p - PQ and D 1(modp), • ⌘ then p up+1. P 2 4 | Maple’sFind Version first P where theSuppose Jacobin symbol7(mod12)is = 1. • ⌘ n Idea: Given a large positiveprime. integer We cann,if taken✓isP prime,=4.◆ there is Should check if n is a square (recall(p 10931)/2 and 3511). a 50-50Take• Q chance=1. that a D will satisfy D 1(modn). • TakeMakeQ =1. use of the identities, where n 1. ⌘ •Play with P and Q until you find such a 2D with n - PQ. 2 • v =4,v =14 P,...4 vn+1 = v n 2(n 0) Find first P where the Jacobi1 symbol2 = 2 1. 2 •ComputeSupposeunn+1 7(mod4)isquickly and check if n un+1. Ifn not, then n is • ⌘ | ✓ n ◆ composite.prime. We2 If can so, take thenP it=4. is likely n is prime.✓ ◆ vShould= v check2,v if n is a= squarev v (recallP, 1093 Du and=2Lv 3511).= v Pvn • 2n n 2n+1 n+1 n n nn+1 2 n Make use of the identities, where n 1. •How do we compute un+1 quickly? 2 v1 =4,v2 =14,...p v2n+1 = v2n 2(n 0) N =2 1 is a prime v(N+1)/4 is divisible by N Why does p u if p is a prime? () | p+1 2 v2n = vn 2,v2n+1 = vn+1vn P, Dun =2vn+1 Pvn Why2n shouldn we2 thinkn+1 n nis+1 likelyn a primen if n unn+1+1? n | O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas-LehmerThe Primality Lucas Primality Test Test O ce Hours and Such: Fix integers P andTheQ Lucas-Lehmer.LetD = P 2 Test.4Q. DefineLet p recursivelybe an odd prime, and define Iun willand notvn haveby regularrecursively o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know u0 =0,u1 =1,un+1 = Pun Qun 12for n 1, p if you plan to comeL0 =4 by. and Ln+1 = L 2 (mod (2 1)) for n 0. n v0 =2,v1 = P, andp vn+1 = Pvn Qvn 1 for n 1. Then 2 1 is a prime if and only if Lp 2 =0. Mersenne Primes (p 1)/2 If p is an odd prime and p - PQ and D 1(modp), ⌘ then p u . The Lucas| p+1 Primality Test TheIdea: Lucas-Lehmer Given a large Test. positiveLet integerp be ann,if oddn prime,is prime, and there define is (p 1)/2 recursivelya 50-50 chance that a D will satisfy D 1(modn). ⌘ Play with P and Q until you2 find suchp a D with n - PQ. L0 =4 and Ln+1 = Ln 2 mod (2 1) for n 0. ComputeSupposeunn+1 7(mod4)isquickly and check if n un+1 . If not, then n is p ⌘ | Thencomposite.prime.2 We1 If canis so, a take thenprimeP it=4. if is and likely onlyn is if prime.Lp 2 =0.
How do we compute un+1 quickly? 2 v =4,v =14,... vn+1 = v 2(n 0) 1 2 2 2n Why does p u if p is a prime? | p+1 Why should we think n is likely a prime if n u ? | n+1 O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas Primality Test
The Lucas-Lehmer Test. Let p be an odd prime, and define recursively Suppose n 7(mod4)is2 p LSuppose0 =4 nand7(mod4)is⌘Ln+1 = Ln 2 mod (2 1) for n 0. Supposeprime.n ⌘ We7(mod4)is can take P =4. prime.p We can⌘ take P =4. Thenprime.2 We1 canis a take primeP =4. if and only if Lp 2 =0. 2 v1 =4,v2 =14,... v2n+1 =2 v2n 2(n 0) v1 =4,v2 =14,... v2n+1 = v22n 2( n 0) v1 =4,v2 =14,... v2n+1 = v n 2(n 0) Compute un modulo p by using2 Compute un modulo p by using Ln = v2n un+1 vn+1 Lnn= 1v2nP P Q un+1 vn+1 n =1 MP where MP =Q . un =vMn 02where M = 10. un p✓vn ◆ 02✓ ◆ 10✓ ◆ N✓ =2p 1◆ is a prime✓ ◆ v(N+1)/4 is divisible✓ by ◆N m =2 1 is a prime ()v(m+1)/4 is divisible by m n ()n n n↵ n n ↵un = and nvn = n↵ + for n 0, un = ↵ and vn = ↵ + for n 0, ↵ The Lucaswhere Primality↵ =(P Test+ pD)/2 and =(P pD)/2 where ↵ =(P + pD)/2 and =(P pD) /2 2 Fix integers P nand1 Q.Letn D =n 1P n4Q.n Define3 recursivelyn n 5 2 n 1 2 unn =n 1 P n +n 3 P nD +n 5 2P D + un and2vn byun = P 1+ P 3D + P 5D + ··· 1 ✓ ◆ 3 ✓ ◆ 5 ✓ ◆ ··· ✓ ◆ ✓ ◆ ✓ ◆ u0 =0,u1 =1,un+1 = Pun Qun 1 for n 1, = p up+1 = p u)p+1 | v0 =2),v|1 = P, and vn+1 = Pvn Qvn 1 for n 1. (p 1)/2 If p is an odd prime and p - PQ and D 1(modp), ⌘ then p u . | p+1 Idea: Given a large positive integer n,ifn is prime, there is a 50-50 chance that a D will satisfy D(p 1)/2 1(modn). ⌘ Play with P and Q until you find such a D with n - PQ. Compute u quickly and check if n u . If not, then n is n+1 | n+1 composite. If so, then it is likely n is prime.
How do we compute un+1 quickly? Why does p u if p is an odd prime? | p+1 Why should we think n is likely a prime if n u ? | n+1 O ce Hours and Such:
I will not have regular o ce hours tomorrow. I “should” be around tomorrow afternoon (2:00-4:00), but do let me know if you plan to come by.
Mersenne Primes
The Lucas Primality Test
The Lucas-LehmerSuppose n Test.7(mod4)isLet p be an odd prime, and define ⌘ recursivelyprime. We can take P =4. L =4 and L = L2 2 mod (2p 1) for n 0. Suppose0 n 7(mod4)isn+1 n 2 p v1 =4⌘ ,v2 =14,... v2n+1 = v n 2(n 0) Thenprime.2 We1 canis a take primeP =4. if and only if Lp 22 =0 . Compute un modulo p by using Compute un modulo p by using u v Ln1=Pv2n 2 P Q v =4,vn+1 =14n+1 ,...n vn+1 = v n 2(n 0) n Definition.un+11 vn A+1Mersenne2 n =1 primeMP is2 a prime2where of theMP form =Q 2 1. . un =vMn 02where M = 10 . un ✓vn ◆ 02✓ ◆ 10✓ ◆ N✓ =2p 1◆ is a prime✓ ◆ v is divisible✓ by ◆N Equivalently, ... of the() form( 2Np+1)/14 where p is a prime. • n n n n↵ n n Mersenne↵un primes= are relatedand to nvperfectn = n↵ numbers+ for. Eulern 0, • un = ↵ and vn = ↵ + for n 0, showed that↵ (m )=2m, where m is even if and only if The Lucas Primalityp 1 p Test p m =2where (2↵ =(P1)+ wherepD)/p2and and 2 =(1P are primes.pD)/2 where ↵ =(P + p D)/2 and =(P pD) /2 57885161 Fix integersTheP largestand Q known.LetD prime= P is2 2 4Q. Define1. recursively • n 1 n n 1 n n 3 n n 5 2 u and vn 1by 2 unn =n 1 P n +n 3 P nD +n 5 2P D + n 2 n u = P + P D + P D + (2 +n p3)n (2 1 p3)n 3 5 ··· 1 ✓ ◆ 3 ✓ ◆ 5 p✓ n◆ ···p n uun0==0,u1✓=1◆ ,u n+1 ✓= Pu◆ andn Quvn n=(2+✓1 for◆ n3) +(21, 3) p12 = p up+1 v0 =2= ,vp 1u)=p+1P, | and vn+1 = Pvn Qvn 1 for n 1. ) | (p 1)/2 If p is an odd prime and p - PQ and D 1(modp), ⌘ then p u . | p+1 Idea: Given a large positive integer n,ifn is prime, there is a 50-50 chance that a D will satisfy D(p 1)/2 1(modn). ⌘ Play with P and Q until you find such a D with n - PQ. Compute u quickly and check if n u . If not, then n is n+1 | n+1 composite. If so, then it is likely n is prime.
How do we compute un+1 quickly? Why does p u if p is an odd prime? | p+1 Why should we think n is likely a prime if n u ? | n+1 Suppose n 7(mod4)is ⌘ Definition.prime. A WeMersenne can take primeP =4.is a prime of the form 2n 1. Suppose n 7(mod4)is p Equivalently,⌘ ... of the form 2 1 where2 p is a prime. prime. Wev1 =4 can,v take2 =14P =4.,... vn+1 = v n 2(n 0) • 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L where= v nm is2 even if and only if v1 =4,v2 =14,...n v2n+1 2= v2n 2(n 0) m =2p 1(2p 1) where p and 2p 1 are primes. p NThe=2 largest1 is known a prime prime isv 2(57885161N+1)/4 is divisible1. by N • () (2 + p3)n (2 p3)n u( ==): and v =(2+p3)n+(2 p3)n ( (n =):(=): p n (( 12 (= ): (=(=)): ): )) (N 1)/2 (N 1)/2 3(N 1)/2 1(modN) and 2(N 1)/2 1(modN) • ⌘ ⌘ It su ces to prove v(N+1)/2 2(modN). • (N+1)/2 ⌘ 2 2 p3=((p2 p6)/2)2 • ± ±
• N+1 N+1 p2+p6 p2 p6 v(N+1)/2 = + v(N+1)/2 = 2 + 2 2 ! 2 !
(N+1)/2 N N +1 N+1 2j 2j =2 N p2 p6 =2 2j 2 6 j=0 ✓ 2j ◆ Xj=0 ✓ ◆ (N+1)/2 (1 N)/2 N +1 j =2(1 N)/2 3j =2 2j 3 j=0 ✓ 2j ◆ Xj=0 ✓ ◆
(N 1)/2 (N+1)/2 2(N 1)/2v(N+1)/2 1+3(N+1)/2 2(modN) • (N+1)/2 ⌘ ⌘ Suppose n 7(mod4)is ⌘ Definition.prime. A WeMersenne can take primeP =4.is a prime of the form 2n 1. Suppose n 7(mod4)is p Equivalently,⌘ ... of the form 2 1 where2 p is a prime. prime. Wev1 =4 can,v take2 =14P =4.,... vn+1 = v n 2(n 0) • 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L where= v nm is2 even if and only if v1 =4,v2 =14,...n v2n+1 2= v2n 2(n 0) m =2p 1(2p 1) where p and 2p 1 are primes. p NThe=2 largest1 is known a prime prime isv 2(57885161N+1)/4 is divisible1. by N • () (2 + p3)n (2 p3)n u( ==): and v =(2+p3)n+(2 p3)n ( (n =):(=): p n (( 12 (= ): (=(=)): ): )) (N 1)/2 (N 1)/2 3(N 1)/2 1(modN) and 2(N 1)/2 1(modN) • 3 ⌘ 1(modN) and 2 ⌘ 1(modN) • ⌘ ⌘ It su ces to prove v(N+1)/2 2(modN). • It su ces to prove v(N+1)/2 ⌘ 2(modN). • ⌘ 2 2 p3=((p2 p6)/2)2 • 2 ± p3=((p2 ± p6)/2) • ± ±
• • N+1 N+1 p2+p6 N+1 p2 p6 N+1 v = p2+p6 + p2 p6 v(N+1)/2 = + v(N+1)/2 = 2 ! + 2 ! 2 ! 2 ! (N+1)/2 (N+1)/2 N +1 N+1 2j 2j N N +1 p N+1 2jp 2j =2 N p2 p6 =2 2j p2 p6 j=0 ✓ 2j ◆ Xj=0 ✓ ◆ X ✓ ◆ (N+1)/2 (N+1)/2 N +1 (1 N)/2 N +1 j =2(1 N)/2 3j =2 2j 3 j=0 ✓ 2j ◆ Xj=0 ✓ ◆ X ✓ ◆
(N 1)/2 (N+1)/2 2(N 1)/2v(N+1)/2 1+3(N+1)/2 2(modN) • 2 v(N+1)/2 ⌘ 1+3 ⌘ 2(modN) • ⌘ ⌘ Suppose n 7(mod4)is ⌘ Definition.prime. A WeMersenne can take primeP =4.is a prime of the form 2n 1. Suppose n 7(mod4)is p Equivalently,⌘ ... of the form 2 1 where2 p is a prime. prime. Wev1 =4 can,v take2 =14P =4.,... vn+1 = v n 2(n 0) • 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L where= v nm is2 even if and only if v1 =4,v2 =14,...n v2n+1 2= v2n 2(n 0) m =2p 1(2p 1) where p and 2p 1 are primes. p NThe=2 largest1 is known a prime prime isv 2(57885161N+1)/4 is divisible1. by N • () (( =):=): (( (2 + p3)n (2 p3)n ( =): p n p n u(n (==): and vn =(2+ 3) +(2 3) (=(=( ( =):():):=): p12 ())( (= ): (=((NN ):1)1)//22 ((NN 1)1)//22 (=(=33)): ): 1(mod1(modNN)) and and 2 2 1(mod1(modNN)) ••)) ⌘ ⌘ ⌘⌘ ) (N 1)/2 (N 1)/2 3(N 1)/2 1(modN) and 2(N 1)/2 1(modN) • 3(N 1)/2 ⌘ 1(modN) and 2(N 1)/2 ⌘ 1(modN) ItIt• su su3 cesces to to⌘ prove prove1(modvv((NN+1)+1)N//)22 and2(mod 22(mod N⌘N).).1(modN) •• • ⌘ ⌘ ⌘ ⌘ It su ces to prove v(N+1)/2 2(modN). • Itpp su cespp to provepp v(N22+1)/2 ⌘ 2(modN). 22• It su3=((3=(( ces to22 prove6)6)//v2)2)(N+1)/2 ⌘ 2(modN). •• •±± ±± ⌘ p p p 2 2 p3=((p2 p6)/2)2 • 2 ± p3=((p2 ± p6)/2)2 •• • 2 ± 3=(( 2 ± 6)/2) • ± ± NN+1+1 NN+1+1 pp pp pp pp • 2+2+ 66 22 66 • vv((NN+1)+1)//22 == N+++1 N+1 • p222+p6 N+1 p222 p6 N+1 p2+p!!6 N+1 p2 p!!6 N+1 v(N+1)/2 = p p + p p v(N+1)/2 = 2+ 6 + 2 6 ((NN+1)+1)2//22 ! 2 ! v(N+1)/2 = 2 ! + 2 ! NN 2 NN!+1+1 NN+1+1 222jj 2!2jj =2=2 pp22 pp66 (N+1)/2 (N+1)/2 22Njj +1 N+1 2j 2j Njj=0=0(N+1)✓✓/2 N +1◆◆ p N+1 2jp 2j =2 NXX N +1 p2N+1 2jp62j =2 N 2j p2 p6 =2 ((NNj=0+1)+1)//22✓ 2j ◆ 2 6 Xj=0 ✓ NN2j+1+1◆ (1(1 NN))//22 Xj=0 ✓ ◆ jj =2=2 X(N+1)/2 33 (N+1)/2 22Njj +1 (1 N)/2jj=0=0(N+1)✓✓/2 N +1◆◆ j =2(1 N)/2XX N +1 3j =2(1 N)/2 2j 3j =2 j=0 ✓ 2j ◆3 Xj=0 ✓ 2j ◆ ((NN 1)1)//22 ((NN+1)+1)Xj=0//22 ✓ ◆ 22 vv 1+31+3 X ✓ 2(mod2(mod◆ NN)) •• ((NN+1)+1)//22 ⌘⌘ X ⌘ ⌘ (N 1)/2 (N+1)/2 2(N 1)/2v(N+1)/2 1+3(N+1)/2 2(modN) • 2(N 1)/2v(N+1)/2 ⌘ 1+3(N+1)/2 ⌘ 2(modN) • 2 v(N+1)/2 ⌘ 1+3 ⌘ 2(modN) • ⌘ ⌘ Suppose n 7(mod4)is ⌘ prime. We can take P =4.
2 Supposevn=47(mod4)is,v =14,... vn+1 = v n 2(n 0) 1 ⌘ 2 2 2 ((( prime.=):=):=): We can take P =4. ((( Ln = v2n (=(=(=( ):):):=): 2 ( =):v =4,v =14,... vn+1 = v 2(n 0) )())((=):1 2 2 2n ( =):( p (N(((NNN=21)1)1)///222 1 is a prime v(((NNN +1)1)1)1)///2224 is divisible by N (=(=333 ):): 1(mod1(mod1(modNNN)))() and and and 2 2 2 1(mod1(mod1(modNNN))) •••(=)) ): ⌘ ⌘ ⌘ ⌘⌘⌘ (= )): ) (N 1)/2 (N 1)/2 ItItIt su su su3( N cescesces1)/2 to to to prove prove prove1(modvvv N) and2(mod 22(mod2(mod(N 1)/2 NNN).).).1(modN) • 3(N 1)/2 ⌘ 1(mod(((NNN+1)+1)+1)N///)222 and 2(N 1)/2 ⌘ 1(modN) ••• • 3 ⌘ 1(modN) and⌘ ⌘ ⌘ 2 ⌘ 1(modN) • ⌘ ⌘ 222 Itppp su3=((3=((3=(( cesppp to222 proveppp6)6)6)///v2)2)2)(N222+1)/2 2(modN). • It su ces to prove v(N+1)/2 ⌘ 2(modN). ••• •±±±It su ces to±±± prove v(N+1)/2 ⌘ 2(modN). • ⌘ 2 2 pp3=((pp2 pp6)/2)2 ••• • 2 ± 3=(( 2 ± 6)/2)2 • 2 ± p3=((p2 ± p6)/NN2)N+1+1+1 NNN+1+1+1 • ± ppp2+2+2+± ppp666 ppp222 ppp666 • vvv(((NNN+1)+1)+1)///222 === +++ • 222 N+1 222 N+1 • p p!!! N+1 p p!!! N+1 p2+2+p66 N+1 p22 p66 N+1 v(N+1)/2 = p p + p p v(N+1)/2 = (((NNN+1)+1)2++1)///222 6 + 2 6 2 NN+1+1 222jj 22jj v(N+1)/2 =NNN 2 NNN!!+1+1+1+ N+1 22j !!2j =2=2=2 2 ! ppp 222 2 ppp666! !22jj ! jj=0=0(N+1)/2 2j j=0(N+1)✓✓✓/2 N +1◆◆◆ N+1 2j 2j NXXX(N+1)/2 N +1 N+1 2j 2j =2 N N +1 p2N+1 2jp62j =2 N +1 p2N+1 2j p62j =2 N (((NNN+1)+1)+1)///222 2j p2 p6 =2 j=0 ✓ NNN2j+1+1+1◆ 2 6 (1(1(1 NNN)))///222 Xj=0 ✓ 2j ◆ jjj =2=2=2 Xj=0 ✓ ◆ 333 X(N+1)/2 222jjj jjj=0=0=0(N+1)✓✓✓/2 N +1◆◆◆ (1 N)/2XXX(N+1)/2 N +1 j =2(1 N)/2 N +1 3j =2(1 N)/2 2j 3j =2 j=0 ✓ 2j ◆3 (((NNN 1)1)1)///222 (((NNN+1)+1)+1)Xj=0///222 ✓ 2j ◆ 222 vvv(((NNN+1)+1)+1)///222 1+31+31+3 Xj=0 2(mod2(mod2(modNNN))) ••• ⌘⌘⌘ X ⌘ ⌘ ⌘ ✓ ◆ (N 1)/2 (N+1)/2 2(N 1)/2v(N+1)/2 1+3(N+1)/2 2(modN) • 2(N 1)/2v(N+1)/2 ⌘ 1+3(N+1)/2 ⌘ 2(modN) • 2 v(N+1)/2 ⌘ 1+3 ⌘ 2(modN) • ⌘ ⌘ Suppose n 7(mod4)is ( (=):=): ⌘ ( ( prime. We can take P =4. (=(=): ): )Suppose) n 7(mod4)is 2 v1 =4⌘ ,v2 =14,... v2n+1 = v n 2(n 0) prime.(N We1)/ can2 take P =4. (N 1)/22 3(N3 1)/ 2 1(mod1(modN)N and) and 2(N 2 1)/ 2 1(mod1(modN)N) • • ⌘ ⌘ ⌘ ⌘ n Ln = v2 2 It suItv1 su=4ces ces,v to2 to prove=14 prove,...v(Nv+1)(N/+1)2 v/22n+12(mod=2(modv2n N2().Nn). 0) ( •=):• ⌘ ⌘ ( p ( ( =):N=):=2p 1 is ap primep 2 2 v(N+1)/4 is divisible by N ((2 2p3=(( 3=((p2 2p6)/6)2)()/2) (=• (•):± ± ± ± () =): (=(=(): ): •)•()N 1)/2 (N 1)/2 (=3 ): 1(modN) andN+1N 2 +1 1(modN+1NN+1) • ) (N 1)⌘ /2 p2+p2+p6p6 (Np21)p⌘/22p6p6 3 1(modN) and 2 1(modN) • v(Nv+1)(N/+1)2 =⌘ /2 = + + ⌘ • (N 1)/2 ⌘ 2 (N 1)/2 ⌘2 It su3 ces to prove 1(mod 2v(N+1)N!/)2 ! and2(mod 2 2 N).1(mod! ! N) • • ⌘ ⌘ ⌘ It su ces to prove v(N+1)/2 2(modN). • (N+1)(N/+1)2 2/2 ⌘ 2 p3=((p2 p6)/2) N +1N +1 N+1N+12j 2j 2j 2j It su ces to proveN N v(N+1)/2 2(modp N ).p • •± p =2=2 p± p 2 ⌘ p2 2 p6 6 2 p3=((p2 p6)/2)2j 2j • ± ±j=0j=0 ✓ ◆ p p XpX✓ 2 ◆ • 2 3=(( 2 6)/2) • ± ± (NN+1)+1/2 N+1 • p p(N+1)/2 Np+1 p • (1 2+N(1)/N2 )/62 NN+1 +12 j j6 N+1 v(N+1)/2 ==2=2 p p N++1 p3 3 p N+1 • p22+p6 2j 2j p22 p6 v(N+1)/2 = j=0! j=0N+1✓ + ◆ ! N+1 v(N+1)/2 = p XpX✓ + ◆p p 2+2 6! 2 2 6! v(N+1)/2 = (N+1)/2 + (N 1)/2 N 2 N(N+1)+1/2 N+1 22j 2j (N2 1)/ 2 v (N1+3+1)(N/+1)2!/2 p 2(mod p N! ) 2 v(N=2+1)(N/+1) 2 /2 1+3 N +12(mod2 N+1N26j) 2j • • ⌘ N⌘ 2Nj ⌘ +1⌘ p N+1 2jp 2j =2 j=0(N+1)✓/2 ◆ p2 p6 X N2+1j N+1 2j 2j N j=0 ✓ ◆p p =2 (NX+1)/2✓ ◆ 2 6 X N2j+1 (1 N)/2 j=0(N+1)✓/2 ◆ j =2 X(N+1)/2 N +13 (1 N)/2 2Nj +1 j =2 j=0(N+1)✓/2 ◆ 3 X N2+1j (1 N)/2 j=0 ✓ 2j ◆ j =2 Xj=0 ✓ ◆3 X 2j (N 1)/2 (N+1)j=0/2 ✓ ◆ 2 v(N+1)/2 1+3 X 2(modN) • (N 1)/2 ⌘ (N+1)⌘ /2 2(N 1)/2v(N+1)/2 1+3(N+1)/2 2(modN) • (N+1)/2 ⌘ ⌘ 2(N 1)/2v 1+3(N+1)/2 2(modN) • (N+1)/2 ⌘ ⌘ Suppose n 7(mod4)is ⌘ prime. We can take P =4.
Suppose n 7(mod4)is 2 v1 =4⌘ ,v2 =14,... vn+1 = v n 2(n 0) prime. We can take P =4. 2 2
n Ln = v2 2 v =4,v =14,... vn+1 = v 2(n 0) 1 2 2 2n p ( =):N =2 1 is a prime v(N+1)/4 is divisible by N ( () (=( ==):): Note that this is the important direction! () (= ): ) (N 1)/2 (N 1)/2 3 1(modN) and 2 1(modN) • ⌘ ⌘ It su ces to prove v 2(modN). • (N+1)/2 ⌘ 2 p3=((p2 p6)/2)2 • ± ±
• N+1 N+1 p2+p6 p2 p6 v(N+1)/2 = + 2 ! 2 !
(N+1)/2 N N +1 N+1 2j 2j =2 p2 p6 2j Xj=0 ✓ ◆ (N+1)/2 (1 N)/2 N +1 j =2 3 2j Xj=0 ✓ ◆
(N 1)/2 (N+1)/2 2 v 1+3 2(modN) • (N+1)/2 ⌘ ⌘ Suppose n 7(mod4)is Definition. A Mersenne⌘ prime is a prime of the form 2n 1. prime. We can take P =4. p SupposeEquivalently,n 7(mod4)is... of the form 2 1 where2 p is a prime. • v1 =4⌘ ,v2 =14,... vn+1 = v n 2(n 0) prime. We can take P =4. 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L wheren = v2nm is even if and only if p 1 p p 2 mv=21 =4,v(22 =141),... where p vand2n+1 2= v21n are2( primes.n 0) NThe=2 largestp 1 is known a prime prime isv 257885161 is divisible1. by N • () (N+1)/4 p n p n (2 + 3) (2 3) n n un = and vn =(2+p3) +(2 p3) p12
( ==): ): Note that this is the important direction! ( 2 (2 p3)2 1= p12 (2 p3) (all signs the same) •(2 ±p3)2 1= ±p12 (2 ±p3) (all signs the same) (=• (2±): p3)2 1=±p12 (2±p3) (all signs the same) • )(2±p3) 1=±p12 (2±p3) (all signs the same) • uv0±n=0= ,uun+1 1 =1u±n,u1 andn+1±u=mPu+n =n umQuunn+11 =4umun1unun 1 •vn(N= u1)n/+12 un 1 and um+n = (uN mu1)n/+12 um 1u n • v3n = un+1 u1(modn 1 andNu)m and+n = 2um un+1 1(modum 1unN) • vn = uvn+10 =2 u,vn 1 1 and= P u=4m+n,v= unm+1u=4n+1 v n umv n1u1n •• e ⌘ ⌘ If ep un with e 1, then Future Homework • If pe uDu|n with=2ev 1, thenPv = 6u = v 2v • IfItp sue| unceswithn to provee n +11,v then(N+1)n /2 2(modn n+1N). n •• If p |unkwith1 e 1, thene+1 )⌘ k e+1 • ukn |kuk n1 +1un (mod ep+1 ) and ukn+1 ku (mod ep+1 ). u ku u (mod p ) and u u n+1(mod p ). kn ⌘ nk+11 n e+1 2 kn+1 ⌘ nk+1 e+1 ukn2⌘ kupkn3=(( +11 un p(mod2 ppe6)+177232917/)2) and ukn+1 ⌘ ukn+1 (mod pe+1). ukn ⌘ ku un (mod p 2 ) and ukn1 +1 ⌘ u (mod p ). • ⌘± nBEWARE+1 ± BAD NOTATION⌘ n+1 u•(k+1)n = uknun+1 ukn 1un = uknun+1 + un(ukn+1 4ukn) u(k+1)n = uknun+1 ukn 1 un p==Nu+1knp un+1 + un(ukn+1N+1 4ukn) u(k+1)n = uknun+1 pukn2+ 1upn6=6 uknun+1p+2un(pukn6 +1 4ukn) u(k+1)n = uknun+1 ukn 1un = uknun+1 + u n(ukn+1 4ukn) v(eN+1)/2 = e+1+ If ep un with e 1,2 then!ep+1 upn . 2 ! •If pe u|n with e 1, then pe+1 u|pn. • If pe|un with e 1, then pe+1|upn. • If p |un with e 1,(N then+1)/2 p |upn. • primes| p, " = "p N1,|+10, 1 suchN+1 that2j p u2jp+". primes p, " =N" 1, 0, 1 such that p u . • 8primes p, =29" = "p 2 { 1, 0, 1 }suchp2 that ppu6|p+". • 8 primes p, 9 " = "p 2 { 1, 02,j1} such that p|up+". • 8 9 p j2=0{ ✓ } ◆ | p+" • u80 =0,u1 =19 ,u2X2=4{ ,u3 =15} ,... ("2 |= "3 =0) u0 =0,u1 =1,u2 =4,u3 =15,... ("2 = "3 =0) u0 =0,u1 =1,u2 =4(N,u+1)/23 =15,... ("2 = "3 =0) u0 =0,u1 =1,u2 =4,u3 =15,... ("2 = "3 =0) (1 N)/2 N +1 j n 1 =2 3 n 1 2 n/2 b c n 2j n/b 2 c n b n2 1c n 2k j1=0k ✓ b ◆c n 2k+1 k un =n 2 1 n 2 X3 ,vn =n/2 n 2 3 b 2 c n n 2k 1 k bn/2c n n 2k+1 k un = b c 2nk +12n 2k 13k,vn = b c n2k2n 2k+13k un = k=0 ✓2k +1 ◆2n 2k 13k,vn = k=0 ✓2k ◆2n 2k+13k un =(Nk=0X1)/✓2 2k +1◆2 3(N,v+1)/2n = k=0X✓2k◆2 3 2 Xk =0 ✓v2k +1◆ 1+3 Xk2(mod=0 ✓2k◆N) kX=0 (N+1)/2 kX=0 • X ✓(p 1)/2 ◆⌘ ⌘ X ✓ ◆ up (3p 1) /2 1(modp) and vp 4(modp) up ⌘3(p 1)/2 ⌘ ±1(modp) and vp ⌘4(modp) up ⌘ 3(p 1)/2 ⌘ ±1(modp) and vp ⌘ 4(modp) up ⌘ 3 ⌘ ±1(modp) and vp ⌘ 4(modp) ⌘ ⌘ ± ⌘ sfds sfds sfds sfds Suppose n 7(mod4)is Definition. A Mersenne⌘ prime is a prime of the form 2n 1. prime. We can take P =4. p SupposeEquivalently,n 7(mod4)is... of the form 2 1 where2 p is a prime. • v1 =4⌘ ,v2 =14,... vn+1 = v n 2(n 0) prime. We can take P =4. 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L wheren = v2nm is even if and only if p 1 p p 2 mv=21 =4,v(22 =141),... where p vand2n+1 2= v21n are2( primes.n 0) NThe=2 largestp 1 is known a prime prime isv 257885161 is divisible1. by N • () (N+1)/4 p n p n (2 + 3) (2 3) n n un = and vn =(2+p3) +(2 p3) p12 (2 p3)2 1= p12 (2 p3) (all signs the same) ( ==):• ): Note± that this± is the important± direction! ( 2 (2 p3)2 1= p12 (2 p3) (all signs the same) •(2 v±np=3)u2 n+1 1=un±p112and (2 ±upm3)+n (all= u signsmun+1 the same)um 1un (=• (2• ±): p3)2 1= ±p 12 (2±p3) (all signs the same) • )(2±p3) 1=±p12 (2±p3) (all signs the same) • v±n = ue n+1 u±n 1 and ±um+n = umun+1 um 1un If p un with e 1, then •vn(N= u1)n/+12 un 1 and um+n = (uNmu1)n/+12 um 1un • v•3n = un+1| u1(modn 1 and Nu)m and+n = 2um un+1 1(modum 1unN) • vn = un+1 un 1 and um+n = umun+1 um 1un • e ⌘ ⌘ • If ep unk with1 e 1, thene+1 k e+1 •uIfknpe u|kun withn +1une (mod 1, thenp ) and ukn+1 un+1 (mod p ). • IfItp sue⌘| unceswith to provee 1,v then(N+1)/2 2(mod⌘N). •• If p |unkwith1 e 1, thene+1 ⌘ k e+1 • ukn |kuk n1 +1un (mod ep+1 ) and ukn+1 ku (mod ep+1 ). u ku u (mod p ) and u u n+1(mod p ). kn ⌘ nk+11 n e+1 2 kn+1 ⌘ nk+1 e+1 ukn2⌘ kupkn3=(( +11 un p(mod2 ppe6)+1/)2) and ukn+1 ⌘ ukn+1 (mod pe+1). ukn ⌘ kun +1un (mod p ) and ukn+1 ⌘ un+1 (mod p ). •u(k⌘+1)±n = uknun+1 ± ukn 1un = uknun+1⌘ + un(ukn+1 4ukn) u•(k+1)n = uknun+1 ukn 1un = uknun+1 + un(ukn+1 4ukn) u(k+1)nIf=pueknuunwith+1 eukn 1,1 u thenn = Nup+1kne+1unu+1 +. un(ukn+1N+1 4ukn) u(k+1)•n = u|knun n+1 pukn 2+ 1upn6= uknun|+1pnp+2un(pukn6 +1 4ukn) u(k+1)n = uknun+1 ukn 1un = uknun+1 + u n(ukn+1 4ukn) v(eN+1)/2 = e+1+ If ep un with e 1,2 then!ep+1 upn . 2 ! •If pe primesu|n withpe, "1,= then"p pe+11,u|0pn, 1. such that p up+". • If• p8e|un with e 9 1, then2p{e +1|upn. } | • If p |un with e 1,(N then+1)/2 p |upn. • primes| p, " = "p N1,|+10, 1 suchN+1 that2j p u2jp+". u0primes=0,up, 1 =1" =,uN" 2 =41,u, 0, 13 =15such,... that (p"2u= ".3 =0) • 8primes p, =29" = "p 2 { 1, 0, 1 }suchp2 that ppu6|p+". • 8 primes p, 9 " = "p 2 { 1, 02,j1} such that p|up+". • 8 9 p j2=0{ ✓ } ◆ | p+" • u80 =0,u1 =19 ,u2X2=4{ ,u3 =15} ,... ("2 |= "3 =0) u =0,un 1 =1,u=4,u=15,... (" = " =0) 0 1 2 3 n/2 2 3 u0 =0,ub 2 1c =1,u2 =4(N,u+1)/23 =15,... ("2 = "3 =0) u0 =0,u1 =1,un 2 =4,u3 =15,...b (c"2 =n "3 =0) (1 N)/n2 2k 1 k N +1 j n 2k+1 k un =n 1 =2 2 3 ,vn = 3 2 3 n 1 2 2k +1 n/2 2k b k=0c ✓ n ◆ 2j n/b 2k=0c ✓n ◆ b n2 1c n 2k j1=0k ✓ b ◆c n 2k+1 k un =n 2 1X n 2 X3 ,vn =n/2Xn 2 3 b 2 c n n 2k 1 k bn/2c n n 2k+1 k un = b c 2nk +12n 2k 13k,vn = b c n2k2n 2k+13k un = k=0 (✓2pk 1)+1/2 ◆2n 2k 13k,vn = k=0 ✓2k ◆2n 2k+13k un =(uNkp=0X1)/✓32 2k +1◆2 1(mod 3(N,v+1)p/)2n = andk=0Xv✓p 2k◆4(mod2 3p) 2 Xk =0⌘ ✓v2k +1⌘◆ ±1+3 Xk2(mod=0 ✓2⌘k◆N) kX=0 (N+1)/2 kX=0 • X ✓(p 1)/2 ◆⌘ ⌘ X ✓ ◆ up (3p 1) /2 1(modp) and vp 4(modp) up ⌘3(p 1)/2 ⌘ ±1(modp) and vp ⌘4(modp) up ⌘ 3(p 1)/2 ⌘ ±1(modpsfds) and vp ⌘ 4(modp) up ⌘ 3 ⌘ ±1(modp) and vp ⌘ 4(modp) ⌘ ⌘ ± ⌘ sfds sfds sfds sfds Suppose n 7(mod4)is Definition. A Mersenne⌘ prime is a prime of the form 2n 1. prime. We can take P =4. p SupposeEquivalently,n 7(mod4)is... of the form 2 1 where2 p is a prime. • v1 =4⌘ ,v2 =14,... vn+1 = v n 2(n 0) prime. We can take P =4. 2 2 Mersenne primes are related to perfect numbers. Euler • showed that (m)=2m,L wheren = v2nm is even if and only if p 1 p p 2 mv=21 =4,v(22 =141),... where p vand2n+1 2= v21n are2( primes.n 0) NThe=2 largestp 1 is known a prime prime isv 257885161 is divisible1. by N • () (N+1)/4 2 (2 p3)2 1= p12 (2 p3) (all signs the same) • (2±(2p +3)p 3)1=n ±(2p12p (23)±n p3) (all signs the same) u• =± ± ± and v =(2+p3)n+(2 p3)n n p n vn = un+1 u12n 1 and um+n = umun+1 um 1un • vn = un+1 un 1 and um+n = umun+1 um 1un • ( ==): ):e Note that this is the important direction! If pe un with e 1, then ( If p un with2 e 1, then • (2 |p3)2 1= p12 (2 p3) (all signs the same) • •(2 ±p| 3)2 1= ±p12 (2 ±p3) (all signs the same) (=• (2±): p3)k 21 1=±p12e+1 (2±p3) (all signsk the same) e+1 u (2 kup3)k 1 u1=(modp12pe+1 (2) andp3)u (all signsuk the(mod same)pe+1). •kn) ± n +1 n ± ± kn+1 n+1 u•kn v⌘±n ku= nu+1n+1 un (modu±n 1 pand)±u andm+n u=knu+1mu⌘n+1un+1u(modm 1unp ). ⌘ ⌘ •vn(N= u1)n/+12 un 1 and um+n = (uNmu1)n/+12 um 1un • v3n = un+1 u1(modn 1 andNu)m and+n = 2um un+1 1(modum 1unN) • vn = un+1 un 1 and um+n = umun+1 um 1un • e ⌘ ⌘ • If ep un with e 1, then • If pe u|n with e 1, then u(•k+1)IfItn p= sue| uknnceswithun to+1 provee ukn1,v1 thenun = uknun2(mod+1 + un(Nukn).+1 4ukn) u(k+1)Ifn p= uknn withun+1 e ukn1, 1 then(uNn+1)=/2uknun+1 + un(ukn+1 4ukn) •• | k 1 e+1 ⌘ k e+1 • ukn |kuk n1 +1un (mod ep+1 ) and ukn+1 ku (mod ep+1 ). u ku u (mod p ) and u u n+1(mod p ). kn ⌘ nk+11 n e+1 2 kn+1 ⌘ nk+1 e+1 uknIf2⌘pekuupkn3=(( +1with1 un p(mode 2 1,pp thene6)+1/)2)p ande+1 uukn.+1 ⌘ ukn+1 (mod pe+1). u•knIf⌘p±ekuunn +1withun (mode ±1,p then)p ande+1 uupnkn.+1 ⌘ un+1 (mod p ). •• ⌘ || n || pn ⌘ primes p, " = "p 1, 0, 1 such that p up+". u•(k+1)primesn = uknpu, n+1" = u"knp 1un 1=, 0u,kn1 usuchn+1 + thatun(upknu+1p+". 4ukn) u(••k+1)88 n = uknun+199 ukn 221 u{{n = Nu+1knu}}n+1 + un(ukn+1|| N+1 4ukn) u(k+1)n = uknun+1 pukn2+ 1upn6= uknun+1p+2un(pukn6 +1 4ukn) u(k+1)n = uknun+1 ukn2+ 1un6= uknun+1 +2un(ukn6 +1 4ukn) u0 =0v,u(N+1)1/=12 =,u 2 =4,u3 =15+ ,... ("2 = " 3 =0) u0 =0,ue 1 =1,u2 =4,ue3+1=15,... ("2 = "3 =0) If ep un with e 1,2 then!ep+1 upn . 2 ! •If pe u|n with e 1, then pe+1 u|pn. • If pe|un with e 1, then pe+1|upn. • If p n|u1n with e 1,(N then+1)/2 p |upn. • primesn| 2 1 p, " = "p N1,|+10, 1 suchn/2N+1 that2j p u2jp+". b 2 c n N bn/2c n • 8primesb c p,n=29" = n"p2k 21 {k 1, 0, 1 }suchbp2 c thatn ppun6|p+2k"+1. k u•n 8=primes p, 9 " =2n" p22k 1{3 k,v1, 0, 1n}=such that p2|un p+2k"+1. 3k u =primes p, " =2 " 3 ,v1, 02,j1 =such that p2u . 3 •n 8 2k +19 p j2=0{ ✓ n } ◆ 2k | p+" • u80 =0k=0,u✓2k1 +1=19 ◆,u2X2=4{ ,u3 =15} k,...=0 ✓2k("◆2 |= "3 =0) u0 =0kX,u=0 ✓1 =1,u◆ 2 =4,u3 =15,...kX=0 ✓("2◆= "3 =0) u0 =0X,u1 =1,u2 =4(N,u+1)/23 =15,...X ("2 = "3 =0) u0 =0,u1 =1,u2 =4,u3 =15,... ("2 = "3 =0) (p 1)/2 (1 N)/2 N +1 j up 3n(p1 1)/2 =2 1(mod p) and vp 3 4(modp) up ⌘n 31 2 ⌘ ±1(modp) and vpn/⌘2 4(modp) ⌘b c ⌘n ± 2j n/b 2 ⌘c n b n2 1c n 2k j1=0k ✓ b ◆c n 2k+1 k un =n 2 1 n 2 X3 ,vn =n/2 n 2 3 b 2 c n n 2k 1 k bn/2c n n 2k+1 k un = b c 2nk +12n 2k 13k,vn = b c n2k2n 2k+13k un = k=0 ✓2k +1 ◆2n 2k 1sfds3k,vn = k=0 ✓2k ◆2n 2k+13k un =(Nk=0X1)/✓2 2k +1◆2 sfds3(N,v+1)/2n = k=0X✓2k◆2 3 2 Xk =0 ✓v2k +1◆ 1+3 Xk2(mod=0 ✓2k◆N) kX=0 (N+1)/2 kX=0 • X ✓(p 1)/2 ◆⌘ ⌘ X ✓ ◆ up (3p 1) /2 1(modp) and vp 4(modp) up ⌘3(p 1)/2 ⌘ ±1(modp) and vp ⌘4(modp) up ⌘ 3(p 1)/2 ⌘ ±1(modp) and vp ⌘ 4(modp) up ⌘ 3 ⌘ ±1(modp) and vp ⌘ 4(modp) ⌘ ⌘ ± ⌘ sfds sfds sfds sfds Suppose n 7(mod4)is ⌘ n Definition.(2prime.p3) A2 WeMersenne1= can takep12 primeP (2=4.pis3) a (all prime signs of the the form same) 2 1. • (2 ±p3)2 1=±p12 (2 ±p3) (all signs the same) • (2±pp3)2 2 1=±pp12 (2±pp3) (all signs the same) •(2(2Equivalently,±p3)3)2 1=1=...p±of1212 the (2 (2 form±p3)3) (all (all 2p signs signs1 where the the same) same)p is a prime. ••Supposev±±n = unn +1 7(mod4)isu±±n 1 and±±um+n = umun+12 um 1un •• vn =vu1n=4+1⌘ ,vu2n =141 and,...um+n = v2nu+1 m=un+1v2n u2(m 1nun 0) prime.• We can take P =4. vMersennev=n =u un+1 primesu un 1and areand relatedu um+=n = tou uperfectumun+1 numbersu um u1u.n Euler •vnn = uenn+1+1 unn 11 and umm++nn = ummunn+1+1 umm 11 unn • • If pe un with e 1, then • showedIf p un that with (em)=21, thenm, where m is even if and only if • e| Ln = v2n • e | p 1 p p 2 IfmIfpv=2ep1u=4unwith,v(2with2 e=14e1)1,,... where1, then thenp vand2n+1 2= v21n are2( primes.n 0) •If p u|nn kwith1 e 1, thene+1 k e+1 • ukn |kukn +11 un (mod pe+1 ) and ukn +1 uk (mod pe+1 ). • ukn |ku un (mod p ) and ukn+1 un+1 (mod p ). ⌘ p nk+11 57885161 ⌘ n+1 NThe=2⌘ largestk 1 1 is known a primee prime+1e+1 isv 2(N+1)/4 is⌘ divisible1.k k by eN+1e+1 u ukn kukuk n1 +1u un(mod(modpep+1)) and andu ukn+1 ukun+1(mod(modpep+1).). uknkn• ⌘kun +1 unn (mod p )() and uknkn+1+1 ⌘un+1 (mod p ). ⌘⌘ n+1 ⌘⌘ n+1 p n p n (2 + 3) (2 3) n n uun(k=+1)np= u2knun +1 p ukn 1unp=anduknuvn+1=(2++ unp(u3)kn+1+(2 4upkn3)) u(k(2+1)n =3)uknu1=pn+1 u12kn (21un =3)ukn (allun signs+1 + u then(u same)kn+1 4ukn) • ± 12± ± u u(k+1)=n =u uknu un+1 u ukn u1u=n =u uknu un+1++u u(nu(ukn+1 4u4ukn) ) u((kk+1)+1)nn = uknknunn+1+1 uknkn 11 unn = uknknunn+1+1 + unn(uknkn+1+1 4uknkn) e e+1 vnIf=peunu+1n withun e1 and1, thenum+npe=+1uumpnu.n+1 um 1u n • • If p |un with e 1, then p |upn. • e| e+1| IfIfpeepu unwithwithe e 1,1, then thenpeep+1+1 u upn. . •If p eu|nn with e 1, then p u|pnpn. • If pprimes| un withp, e " =1," thenp 1,|0, 1 such that p up+". •• • 8 primes|| p, 9 " = "p 2 { 1,|0, 1} such that p|up+". • 8 9 2 { } | primesprimesp,p, " =" =" "p 1,10,,01, 1suchsuch that thatp pu up+. ". • 8primesk 1p, 9" = "pp e+12 {1 , 0, 1 }such thatk p u|pp++"". e+1 •uknu80 =0ku,un +1u1n9=1(mod,up22=4{) ,u and u3 }kn=15+1 ,...un+1("2(mod|= "3 p=0)). • u80⌘=0,u1 9=1,u22=4{ ,u3 }=15⌘,... ("2|= "3 =0) u u=00 =0,u,u=11 =1,u,u=42 =4,u,u=153 =15,...,...("("=2 =" "=0)3 =0) u00 =0,u11 =1,u22 =4,u33 =15,... ("22 = "33 =0) n 1 n 21 n/2 b 2 c n bn/2c n b n 1c n 2k 1 k b c n 2k+1 k u(k+1)u n==n 1u knun+1n ukn2n 1u2kn 1=3ku,vknun+1=+ un(uknn +12n 42ku+1kn3)k n n 1 2 n n/n/2 2 un = b2 c 2 3 ,vn =n/b2 c 2 3 bb 2 cc 2nkn+1 n 2k 1 k b c n2nk n 2k+1 k u = k=0 ✓2nk +1◆n2 2 k 1 k3 ,v=b kc=0 ✓n2k◆n2 2 k+1 k3 un =n kX=0 ✓ 2◆n 2k 13k,vn =n kX=0 ✓ 2◆n 2k+13k un = e X 2k +12 3 ,ve+1 n = X 2k2 3 If p uk=0n with22✓kk +1+1e ◆1, then p upn. k=0 22✓kk ◆ kk=0=0X✓✓(p 1)/2 ◆◆ kk=0=0X✓✓ ◆◆ • upXX| 3(p 1)/2 1(modp)| andXXvp 4(modp) up ⌘ 3 ⌘ ±1(modp) and vp ⌘ 4(modp) ⌘(p (p1)/1)2/2 ⌘ ± ⌘ u uprimesp 3(p3 1) p/,2 " =1(mod"1(modp p1)p,)0, and1 andsuchv vp that4(mod4(modp up+"p.)p) • u8pp ⌘⌘3 9⌘⌘±1(mod± 2 { p) and} vpp ⌘⌘4(mod| p) ⌘ ⌘ ± sfds ⌘ u =0,u=1,u=4,usfds =15,... (" = " =0) 0 1 2 sfds3 2 3 sfdssfds
n 1 n/2 b 2 c b c n n 2k 1 k n n 2k+1 k un = 2 3 ,vn = 2 3 2k +1 2k Xk=0 ✓ ◆ Xk=0 ✓ ◆
(p 1)/2 u 3 1(modp) and v 4(modp) p ⌘ ⌘ ± p ⌘
sfds